\(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors. The point \(A\) has position vector \(6 \mathbf { j } \mathrm {~m}\) relative to an origin \(O\). At time \(t = 0\) a particle \(P\) starts from \(O\) and moves with constant velocity ( \(5 \mathbf { i } + 2 \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\). At the same instant a particle \(Q\) starts from \(A\) and moves with constant velocity \(4 \mathrm { ims } ^ { - 1 }\).
Write down the position vectors of \(P\) and of \(Q\) at time \(t\) seconds.
Show that the distance \(d \mathrm {~m}\) between \(P\) and \(Q\) at time \(t\) seconds is such that
$$d ^ { 2 } = 5 t ^ { 2 } - 24 t + 36 .$$
Find the value of \(t\) for which \(d ^ { 2 }\) is a minimum.
Hence find the minimum distance between \(P\) and \(Q\), and state the position vector of each particle when they are closest together.